Could you please explain why Q=CV?

Discussion in 'Electronic Design' started by Boki, Jul 24, 2003.

  1. Boki

    Boki Guest

    Hi, All:

    Could you please explain why Q=C*V?

    Thanks.

    Boki
     
    Boki, Jul 24, 2003
    #1
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  2. Boki

    John Fields Guest

    On Thu, 24 Jul 2003 10:59:29 +0800, "Boki" <>
    wrote:

    >Hi, All:
    >
    >Could you please explain why Q=C*V?

    ---
    If you have an air compressor with the tank pressure at 100PSI it will
    be filled with twice as many air molecules than if its pressure was at
    50PSI.

    Looking at it another way, if you have an air compressor with a 5 gallon
    tank and another one with a 10 gallon tank and they're both filled to
    100PSI, the 10 gallon tank will have twice as much air in it as the 5
    gallon one.

    C is analogous to the capacity of the tank, Q to the quantity of air
    molecules it's holding, and V to the tank pressure.

    --
    John Fields
     
    John Fields, Jul 24, 2003
    #2
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  3. Boki

    John Fields Guest

    On Thu, 24 Jul 2003 08:07:39 GMT, (John Crighton)
    wrote:


    >
    >Lets say you have a tank with a bottom area of 10 square feet.
    >Lets say that your neighbour in an identical house next door to
    > you buys a rain water tank with a bottom area of 20 square feet.
    >
    >During the night it rains.
    >
    >In the morning, you check your tank and it has 1 foot of water.
    >Your neighbour shouts across the fence "Hey Boki, how much
    >water have you got? I have 6 inches."


    ---
    ???

    Unless it rained twice as hard over the 10 square foot tank as it did
    over the 20 square footer, I think both tanks will have filled to the
    same depth!
    ---

    >You both have the same (Q)uantity of water, 10 cubic feet.
    >(imagine raindrops as electrons)


    ---
    No, the larger tank will have accumulated twice the volume of water as
    the small one.
    ---

    >When you turn your tap on, the water will come out at a
    >faster rate than your neighbour's tap because your tank
    >has a higher head of water.


    ---
    No, both tanks will have the same head.
    ---

    >(Imagine head of water as pressure, think Volts)


    >
    >Q(10 cubic feet of rain drops) = C (area of tank bottom) * V (head
    >of water)
    >
    >Q (quantity of electrons) = C (size of capacitor) * Volts


    ---
    In this case, Q for the large tank comes out to twice what it would be
    for the small tank because for the same head the larger area will allow
    twice the volume of water to be accumulated.

    --
    John Fields
     
    John Fields, Jul 24, 2003
    #3
  4. Boki

    John Fields Guest

    On Thu, 24 Jul 2003 12:39:06 GMT, (John Crighton)
    wrote:


    >Hello John,
    >I should have explained a little better. When I said "both
    >houses are identical" I assumed you would have
    >understood that to mean the catchment area on the
    >rooves (roofs) to be the same.
    >My fault for not explaining that.
    >
    >Back yard rain water tanks are quite common here
    >in Australia. Especially in Adelaide where the water
    >tastes terrible. The tanks are connected
    >to the roof guttering of the house. So with identical
    >houses, meaning identical roofs, you get the same
    >quantity of water.


    ---
    True.
    ---

    >When I said back yard rain water tank I assumed again
    >that the reader would understand that it is for drinking
    >therefore would know it would have a lid on top of the
    >tank. Don't you have backyard rainwater tanks
    >where you live?


    ---
    Perhaps, but I've never seen one. What we have around here is open tanks
    for livestock to drink from.
    ---

    >In Adelaide they are essential.
    >Tea made from tap water is terrible.
    >
    >Oh well, I tried to explain the formula with a water
    >analogy and it did not work. Never mind. :)


    ---
    Actually, it does now!
    ---

    --
    John Fields
     
    John Fields, Jul 24, 2003
    #4
  5. Paul Giampaolo wrote:
    >>Could you please explain why Q=C*V?

    >
    >
    > Farads have the units Coulombs/Volt, therefore total charge measured in
    > coulombs stored on the capacitor is equal to the voltage across the
    > capacitor multiplied by its capacitence.
    >
    > For a plate capacitor, its capacitence can be calaculated by the
    > following equation:
    >
    > C=(A*Er*E0)/D
    >
    > where
    >
    > A is plate area
    > D is distance between plates
    > Er is reltive permittivity of the dielectric
    > E0 is the permittivity of free space
    >
    > E0 has a value of 8.85418781761*10^-12 F/m
    >


    From electrostatic principles you recall (or just accept) that a pair
    of oppositely charged metal plates (i.e. a capacitor) with areas A,
    separated by distance d, with charges +Q and -Q, will have a uniform
    distribution of charges and that the charge density on each plate is Q/A.

    If the plates are close together (i.e. d is small relative to A), then
    the electric field, E, between the plates is constant and is just the
    charge density divided by the permittivity of the dielectric, which is
    roughly e0 for a vacuum or air.

    In other words E = Q/(A*e0).

    Notice that the distance between the plates doesn't appear in the
    equation. This is because they are taken to be so close together that
    the direction of E is perpendicular to the plates at all points
    (neglecting the edges of the plates). Of course this is not the case for
    all capacitors, but it allows us to avoid a lot of calculus and still
    get a result.

    Now we need the the potential difference between the plates, V. Recall
    that V is found by integrating the field E along a path between two
    points in a field and that it is independent of the path taken. It
    represents the energy per charge required to move a charge in a field.
    We can choose the simplest path, which is a straight line perpendicular
    to the plates (in the direction of E) so that,

    V = E*d.

    Plugging in the expression for E we get:

    V = [Q/(A*e0)]*d

    If we rearrange and assign C = A*e0/d, we get

    V = Q/C and Q = C*V

    So what is C? As far as we are concerned here it is the constant of
    proportionality between Q and V. It is a property of the capacitor that
    is directly proportional to the area of the plates and inversely
    proportional to the distance between them, which seems right.

    But why is C inversely proportional to d, when we have already argued
    that the strength of E is independent of d for small values of d? The
    answer is that E is constant, but V is not - it is proportional to d. If
    we were to change d for a given Q (i.e. move the plates), both V and C
    would vary to keep Q = C*V constant.

    Note that the stored energy, 1/2 C*V^2 is not constant with varying d,
    but that's the subject of another lecture.

    --
    Joe Legris
     
    Joseph Legris, Jul 24, 2003
    #5
  6. Boki

    Genome Guest

    "Boki" <> wrote in message
    news:bfni2i$...
    > Hi, All:
    >
    > Could you please explain why Q=C*V?
    >
    > Thanks.
    >
    > Boki
    >
    >


    Before everyone else on the planet loses sight of things I'd like to point
    out that Q=C*V because...... that's the way it's defined.

    Some old dead bloke did some experiments and discovered a relationship
    between what he had and what he got. Later on some other old dead blokes
    stuck their heads together and took what all the other old dead blokes doing
    experiments had discovered and sorted out the units so it all made sense.

    See..... Quite simple really.

    DNA


    ---
    Outgoing mail is certified Virus Free.
    Checked by AVG anti-virus system (http://www.grisoft.com).
    Version: 6.0.501 / Virus Database: 299 - Release Date: 7/14/03
     
    Genome, Jul 24, 2003
    #6
  7. Boki

    Jim Thompson Guest

    On Thu, 24 Jul 2003 16:00:37 +0100, "Genome"
    <genome@the_point_of_going_to_bed.fuzzything> wrote:

    >
    >"Boki" <> wrote in message
    >news:bfni2i$...
    >> Hi, All:
    >>
    >> Could you please explain why Q=C*V?
    >>
    >> Thanks.
    >>
    >> Boki
    >>
    >>

    >
    >Before everyone else on the planet loses sight of things I'd like to point
    >out that Q=C*V because...... that's the way it's defined.
    >
    >Some old dead bloke did some experiments and discovered a relationship
    >between what he had and what he got. Later on some other old dead blokes
    >stuck their heads together and took what all the other old dead blokes doing
    >experiments had discovered and sorted out the units so it all made sense.
    >
    >See..... Quite simple really.
    >
    >DNA
    >


    ROTFLMAO!

    ...Jim Thompson
    --
    | James E.Thompson, P.E. | mens |
    | Analog Innovations, Inc. | et |
    | Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
    | Phoenix, Arizona Voice:(480)460-2350 | |
    | Jim-T@analog_innovations.com Fax:(480)460-2142 | Brass Rat |
    | http://www.analog-innovations.com | 1962 |

    For proper E-mail replies SWAP "-" and "_"

    Get Lolita Out of Debt... Add Three Inches to Your Mortgage!
     
    Jim Thompson, Jul 24, 2003
    #7
  8. Boki

    Ben Bradley Guest

    In sci.electronics.design, Jim Thompson <Jim-T@analog_innovations.com>
    wrote:

    >On Thu, 24 Jul 2003 16:00:37 +0100, "Genome"
    ><genome@the_point_of_going_to_bed.fuzzything> wrote:
    >
    >>
    >>"Boki" <> wrote in message
    >>news:bfni2i$...
    >>> Hi, All:
    >>>
    >>> Could you please explain why Q=C*V?
    >>>
    >>> Thanks.
    >>>
    >>> Boki
    >>>
    >>>

    >>
    >>Before everyone else on the planet loses sight of things I'd like to point
    >>out that Q=C*V because...... that's the way it's defined.
    >>
    >>Some old dead bloke did some experiments and discovered a relationship
    >>between what he had and what he got. Later on some other old dead blokes
    >>stuck their heads together and took what all the other old dead blokes doing
    >>experiments had discovered and sorted out the units so it all made sense.
    >>
    >>See..... Quite simple really.
    >>
    >>DNA
    >>

    >
    >ROTFLMAO!


    I see these letters in this equation and I wonder what the letters
    could mean (though admittedly other responses mention charge,
    capacitance and voltage).

    Quality equals Character times Veracity?



    > ...Jim Thompson
    >--
    >| James E.Thompson, P.E. | mens |
    >| Analog Innovations, Inc. | et |
    >| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
    >| Phoenix, Arizona Voice:(480)460-2350 | |
    >| Jim-T@analog_innovations.com Fax:(480)460-2142 | Brass Rat |
    >| http://www.analog-innovations.com | 1962 |
    >
    > For proper E-mail replies SWAP "-" and "_"
    >
    > Get Lolita Out of Debt... Add Three Inches to Your Mortgage!
     
    Ben Bradley, Jul 24, 2003
    #8
  9. Boki

    John Hall Guest

    On Thu, 24 Jul 2003 16:00:37 +0100, "Genome"
    <genome@the_point_of_going_to_bed.fuzzything> wrote:

    >Some old dead bloke did some experiments and discovered a relationship
    >between what he had and what he got. Later on some other old dead blokes
    >stuck their heads together and took what all the other old dead blokes doing
    >experiments had discovered and sorted out the units so it all made sense.
    >
    >See..... Quite simple really.
    >


    While I admire your capacity(!) for wit, I'll point out that dead
    blokes don't do experiments.

    --
    John W Hall <>
    Calgary, Alberta, Canada.
    "Helping People Prosper in the Information Age"
     
    John Hall, Jul 24, 2003
    #9
  10. Boki

    Ban Guest


    > "Boki" <> wrote in message
    > news:<bfni2i$>...
    >> Hi, All:
    >>
    >> Could you please explain why Q=C*V?
    >>


    > I know there are many tiles missing, but also your question was very
    > broad...
    >


    I dunno why the water analogy is so popular here- in my eyes it seems to add
    a lot to the confusion expressed in this thread.
    Even a small kid knows this is bulls**t, or else the current would be
    flowing out of the mains receptacles continuously, giving us shocks on the
    feet when a sufficient voltage level is reached. Or maybe it is lighter than
    air and floating on the ceiling? Since it cannot be seen...

    The basic laws of electricity cannot be explained with that stupid water
    analogy.
    1. Current always flows in a closed circle, its flow(electrons/s) is
    identical anywhere at any given moment inside this loop.
    2. Current needs some medium (ie. conductor with free electrons, charge
    carriers, ions...) to flow.
    3. Voltage and current create electric/magnetic fields with unique
    properties. These fields are the cause of capacitive/inductive phenomena
    like a capacitor or a coil...

    Just to name a few.

    I would advise to start reading up on these basic laws and only to proceed
    when understood.

    ciao Ban
     
    Ban, Jul 25, 2003
    #10
  11. Boki

    Pat Ford Guest

    "Ban" <> wrote in message
    news:AT3Ua.18806$...
    >
    > > "Boki" <> wrote in message
    > > news:<bfni2i$>...
    > >> Hi, All:
    > >>
    > >> Could you please explain why Q=C*V?
    > >>

    >
    > > I know there are many tiles missing, but also your question was very
    > > broad...
    > >

    >
    > I dunno why the water analogy is so popular here- in my eyes it seems to

    add
    > a lot to the confusion expressed in this thread.
    > Even a small kid knows this is bulls**t, or else the current would be
    > flowing out of the mains receptacles continuously, giving us shocks on the
    > feet when a sufficient voltage level is reached. Or maybe it is lighter

    than
    > air and floating on the ceiling? Since it cannot be seen...
    >
    > The basic laws of electricity cannot be explained with that stupid water
    > analogy.
    > 1. Current always flows in a closed circle, its flow(electrons/s) is
    > identical anywhere at any given moment inside this loop.
    > 2. Current needs some medium (ie. conductor with free electrons, charge
    > carriers, ions...) to flow.
    > 3. Voltage and current create electric/magnetic fields with unique
    > properties. These fields are the cause of capacitive/inductive phenomena
    > like a capacitor or a coil...
    >
    > Just to name a few.
    >
    > I would advise to start reading up on these basic laws and only to proceed
    > when understood.
    >
    > ciao Ban
    >
    >

    think decaf...
    the water analogy is great for some people, there is hole BUT once the
    basics are grasped the fine point can be explained.
    Pat
     
    Pat Ford, Jul 25, 2003
    #11
  12. Boki wrote:
    > Hi, All:
    >
    > Could you please explain why Q=C*V?


    Consider an isolated conducting sphere.

    Now, fire electrons at it. The electrons stick and redistribute
    themselves around the sphere.

    As the charge (ie the number of electrons) on the sphere increases,
    the faster the electrons have to be shot in order to overcome the
    repulsion from the charge on the sphere.

    In essence the electrons have to climb a "potential barrier" in
    order to reach the sphere. This potential is called the voltage
    on the sphere.

    The voltage is proportional to the charge on the sphere.

    The constant of proportionality is called the Capacitance.

    Hence Q=C*V

    It may be easier to understand the alternative formulation
    V=Q/C
    ----------------------------------------------------------------

    Alternatively:-

    Consider two identical capacitors - one uncharged and the other
    with a potential difference of V between its plates.

    Connect the two capacitors in parallel.

    The combined capacitor has twice the capacitance of the two
    individual capacitors.

    The charge remains the same - but is shared between the two
    capacitors. The voltage on the charged capacitor drops to
    one half of its original value.
     
    richard mullens, Jul 25, 2003
    #12
  13. "Jim Thompson" <Jim-T@analog_innovations.com> wrote in message
    news:...
    > On Fri, 25 Jul 2003 12:59:32 -0400, Joseph Legris
    > <> wrote:
    >
    > >richard mullens wrote:
    > >
    > >>
    > >> Alternatively:-
    > >>
    > >> Consider two identical capacitors - one uncharged and the other
    > >> with a potential difference of V between its plates.
    > >>
    > >> Connect the two capacitors in parallel.
    > >>
    > >> The combined capacitor has twice the capacitance of the two
    > >> individual capacitors.
    > >>
    > >> The charge remains the same - but is shared between the two
    > >> capacitors. The voltage on the charged capacitor drops to
    > >> one half of its original value.
    > >>

    > >
    > >But now the energy is 1/2 of what it was in the beginning. Where did it
    > >go?

    >
    > This is going to be fun to watch ;-)
    >
    > (This topic was broached here before, quite a long time ago... the
    > various explanations were hilarious.)




    OK, I'm no physicist, but I'll give it a shot.

    When the two caps are connected, there will be a very large current
    (infinitely large for ideal caps and wiring) flowing through their leads. As
    the leads and wiring are not perfect and the caps' ESRs are > 0, the energy
    will be lost in these non-zero resistances. Some energy will also escape as
    radiation from the spark I guess.

    Costas
     
    Costas Vlachos, Jul 25, 2003
    #13
  14. Boki

    Jim Meyer Guest

    richard mullens <> wrote in message news:<bfrdpi$bko$>...
    >
    > Consider two identical capacitors - one uncharged and the other
    > with a potential difference of V between its plates.
    >
    > Connect the two capacitors in parallel.
    >
    > The combined capacitor has twice the capacitance of the two
    > individual capacitors.
    >
    > The charge remains the same - but is shared between the two
    > capacitors. The voltage on the charged capacitor drops to
    > one half of its original value.


    And since the energy stored in a capacitor is equal to half its
    capacitance multiplied by the square of the voltage across it, after
    you connect the two capacitors together each one will have a *quarter*
    of the energy originally stored in the first capacitor.

    Energy is lost somewhere. Where?

    Jim
     
    Jim Meyer, Jul 25, 2003
    #14
  15. Boki

    mullens Guest

    Jim Thompson wrote:
    >
    > On Fri, 25 Jul 2003 12:59:32 -0400, Joseph Legris
    > <> wrote:
    >
    > >richard mullens wrote:
    > >
    > >>
    > >> Alternatively:-
    > >>
    > >> Consider two identical capacitors - one uncharged and the other
    > >> with a potential difference of V between its plates.
    > >>
    > >> Connect the two capacitors in parallel.
    > >>
    > >> The combined capacitor has twice the capacitance of the two
    > >> individual capacitors.
    > >>
    > >> The charge remains the same - but is shared between the two
    > >> capacitors. The voltage on the charged capacitor drops to
    > >> one half of its original value.
    > >>

    > >
    > >But now the energy is 1/2 of what it was in the beginning. Where did it go?

    >
    > This is going to be fun to watch ;-)
    >


    First of all, we are going to have superconducting wires and plates
    on our capacitors. Then we are going to connect the capacitors using
    a zero RDSon MOSFET - that should get rid of the spark.

    We are left with the sudden flow of electrons through a conductor.
    In this ideal case, the charge will rush out of the charged capacitor
    into the uncharged capacitor. Some energy will be lost in radiated EM
    radiation. Indeed, if one places a compass needle near the wire
    joining the capacitors, some of this energy will be transferred to
    the compass needle when the current is turned on.

    Let us modify the experiment and go back to the isolated conducting
    sphere carrying a charge. Actually this is a charged soap bubble.
    Rather than double the size of the soap bubble by reducing the atmospheric
    pressure (which would double its capacitance - as above), instead
    we halve the capacitance by increasing the pressure.

    Q remains the same, C halves, so V doubles. Hence 1/2 CV2 doubles - so
    now we have twice the electrostatic energy. It is easy to see where
    this extra energy comes from. We are doing work on the system by
    forcing the electrons closer together.
     
    mullens, Jul 26, 2003
    #15
  16. Joseph Legris wrote:
    >
    > richard mullens wrote:
    >
    > >
    > > Alternatively:-
    > >
    > > Consider two identical capacitors - one uncharged and the other
    > > with a potential difference of V between its plates.
    > >
    > > Connect the two capacitors in parallel.
    > >
    > > The combined capacitor has twice the capacitance of the two
    > > individual capacitors.
    > >
    > > The charge remains the same - but is shared between the two
    > > capacitors. The voltage on the charged capacitor drops to
    > > one half of its original value.
    > >

    >
    > But now the energy is 1/2 of what it was in the beginning. Where did it go?
    >
    > --
    > Joe Legris


    Classic em dissertation defense question.

    Analyze charging one capacitor through a resistor R from another. Half
    the energy is always lost.
    Value is independent of R value down to and including R=0.

    In the case of R=0 there is a spark and energy radiates
    electromagnetically.
    Or melts stuff, or welds, or whatever.


    --
    Many thanks,

    Don Lancaster
    Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
    voice: (928)428-4073 email: fax 847-574-1462

    Please visit my GURU's LAIR web site at http://www.tinaja.com
     
    Don Lancaster, Jul 26, 2003
    #16
  17. Boki

    Tweetldee Guest

    "Jim Thompson" <Jim-T@analog_innovations.com> wrote in message
    news:...
    > On Fri, 25 Jul 2003 16:21:21 -0700, Don Lancaster <>
    > wrote:
    >
    > >Joseph Legris wrote:
    > >>
    > >> richard mullens wrote:
    > >>
    > >> >
    > >> > Alternatively:-
    > >> >
    > >> > Consider two identical capacitors - one uncharged and the other
    > >> > with a potential difference of V between its plates.
    > >> >
    > >> > Connect the two capacitors in parallel.
    > >> >
    > >> > The combined capacitor has twice the capacitance of the two
    > >> > individual capacitors.
    > >> >
    > >> > The charge remains the same - but is shared between the two
    > >> > capacitors. The voltage on the charged capacitor drops to
    > >> > one half of its original value.
    > >> >
    > >>
    > >> But now the energy is 1/2 of what it was in the beginning. Where did it

    go?
    > >>
    > >> --
    > >> Joe Legris

    > >
    > >Classic em dissertation defense question.
    > >
    > >Analyze charging one capacitor through a resistor R from another. Half
    > >the energy is always lost.
    > >Value is independent of R value down to and including R=0.
    > >
    > >In the case of R=0 there is a spark and energy radiates
    > >electromagnetically.
    > >Or melts stuff, or welds, or whatever.

    >
    > Don, You done gone and spoiled all our fun. Last time this question
    > was posed we got all kinds of hilarious responses, particularly from
    > those in this crowd who believe the laws of physics may be defied ;-)
    >
    > ...Jim Thompson


    You mean like the EER guy whom I see has done gone and started his tirade up
    again with a new post?
    Heeerre we go again!!!
    --
    Tweetldee
    Tweetldee at att dot net (Just subsitute the appropriate characters in the
    address)

    Time is what keeps everything from happening all at once.
     
    Tweetldee, Jul 26, 2003
    #17
  18. Boki

    Jim Thompson Guest

    On Sat, 26 Jul 2003 03:07:16 GMT, "Tweetldee" <>
    wrote:

    >"Jim Thompson" <Jim-T@analog_innovations.com> wrote in message
    >news:...

    [snip]
    >> Don, You done gone and spoiled all our fun. Last time this question
    >> was posed we got all kinds of hilarious responses, particularly from
    >> those in this crowd who believe the laws of physics may be defied ;-)
    >>
    >> ...Jim Thompson

    >
    >You mean like the EER guy whom I see has done gone and started his tirade up
    >again with a new post?
    >Heeerre we go again!!!


    Must be in my Kill File, I haven't seen that one.

    ...Jim Thompson
    --
    | James E.Thompson, P.E. | mens |
    | Analog Innovations, Inc. | et |
    | Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
    | Phoenix, Arizona Voice:(480)460-2350 | |
    | Jim-T@analog_innovations.com Fax:(480)460-2142 | Brass Rat |
    | http://www.analog-innovations.com | 1962 |

    For proper E-mail replies SWAP "-" and "_"

    Get Lolita Out of Debt... Add Three Inches to Your Mortgage!
     
    Jim Thompson, Jul 26, 2003
    #18
  19. Don Lancaster wrote:
    > Joseph Legris wrote:
    >
    >>richard mullens wrote:
    >>
    >>
    >>>Alternatively:-
    >>>
    >>>Consider two identical capacitors - one uncharged and the other
    >>>with a potential difference of V between its plates.
    >>>
    >>>Connect the two capacitors in parallel.
    >>>
    >>>The combined capacitor has twice the capacitance of the two
    >>>individual capacitors.
    >>>
    >>>The charge remains the same - but is shared between the two
    >>>capacitors. The voltage on the charged capacitor drops to
    >>>one half of its original value.
    >>>

    >>
    >>But now the energy is 1/2 of what it was in the beginning. Where did it go?
    >>
    >>--
    >>Joe Legris

    >
    >
    > Classic em dissertation defense question.
    >
    > Analyze charging one capacitor through a resistor R from another. Half
    > the energy is always lost.
    > Value is independent of R value down to and including R=0.
    >
    > In the case of R=0 there is a spark and energy radiates
    > electromagnetically.
    > Or melts stuff, or welds, or whatever.
    >


    But what about when the capacitors are connected by a Zero RDSon Mosfet
    to eliminate the spark ?
     
    richard mullens, Jul 26, 2003
    #19
  20. Boki

    Bill Bowden Guest

    (Jim Meyer) wrote in message news:<>...
    > richard mullens <> wrote in message news:<bfrdpi$bko$>...
    > >
    > > Consider two identical capacitors - one uncharged and the other
    > > with a potential difference of V between its plates.
    > >
    > > Connect the two capacitors in parallel.
    > >
    > > The combined capacitor has twice the capacitance of the two
    > > individual capacitors.
    > >
    > > The charge remains the same - but is shared between the two
    > > capacitors. The voltage on the charged capacitor drops to
    > > one half of its original value.

    >
    > And since the energy stored in a capacitor is equal to half its
    > capacitance multiplied by the square of the voltage across it, after
    > you connect the two capacitors together each one will have a *quarter*
    > of the energy originally stored in the first capacitor.
    >
    > Energy is lost somewhere. Where?
    >
    > Jim


    I believe half the energy is lost in the resistance connecting the
    capacitors. As resistance goes down, current goes up, and i^2R
    losses stay the same at 50%. To avoid the loss, resistance would
    be zero, and current would be infinite. Is that possible?

    -Bill
     
    Bill Bowden, Jul 26, 2003
    #20
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